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Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of these two vector spaces of cardinalities $c$ and $c\times c = c$, so they are isomorphic as vector spaces over $\mathbb{Q}$.

Is there a way to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups which does not use AC? Are there models of set theory in which these groups are not isomorphic? I'm also curious whether there is a proof (perhaps using AC) which does not make use of this vector space machinery, though I'm fairly sure the above proof is the simplest.

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In addition to the answers below, this is very closely related to this question - mathoverflow.net/questions/16666/… –  François G. Dorais May 20 '10 at 17:12
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up vote 10 down vote accepted

You cannot prove that $\mathbb{R}$ and $\mathbb{R}^{2}$ are isomorphic in $ZF$. To see this, note that the map $(x,y) \mapsto (x,0)$ is a nontrivial non-surjective additive endomorphism of $\mathbb{R}^{2}$. Assuming the existence of suitable large cardinals, every map $f: \mathbb{R} \to \mathbb{R}$ is measurable in $L(\mathbb{R})$ and it follows that the only additive endomorphisms of $\mathbb{R}$ in $L(\mathbb{R})$ are the maps $x \mapsto rx$ for some $r \in \mathbb{R}$. Thus $\mathbb{R} \not \cong \mathbb{R}^{2}$ in $L(\mathbb{R})$.

There is an interesting open problem related to your question of whether an isomorphism can be found which does not make use of the vector space machinery. If we add a Ramsey ultrafilter $\mathcal{U}$ to $L(\mathbb{R})$, then we obtain an interesting model $L(\mathbb{R})[\mathcal{U}]$ which contains the ultrafilter $\mathcal{U}$ while still retaining some of the "nice" properties of $L(\mathbb{R})$. In particular, neither $\mathbb{R}$ nor $\mathbb{R}^{2}$ has a basis as a vector space over $\mathbb{Q}$ in $L(\mathbb{R})[\mathcal{U}]$ . However, it is unknown whether or not $\mathbb{R} \cong \mathbb{R}^{2}$ in $L(\mathbb{R})[\mathcal{U}]$. In other words, does an ultrafilter help in trying to construct such an isomorphism?

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Hi Simon, Have you talked with Paul Larson about this question? With Jindra and Richard, he's been working on properties of the model $L({\mathbb R})[{\mathcal U}]$. I'm sure if he hasn't investigated this, he'll find the problem interesting. –  Andres Caicedo May 20 '10 at 17:07
    
Of course, we have been discussing this kind of problem. They seem quite hard ... –  Simon Thomas May 20 '10 at 17:19
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Here's the way I like to do it:

A group homomorphism with the property of Baire from one Polish group to another is automatically continuous. Of course no bijection between your two groups is a homeomorphism. Solovay provided a model (AC fails) in which every subset of a Polish space has the property of Baire, so in Solovay's model these two groups are not isomorphic.

A more down-to-earth consequence of the result says: no map you can actually write down is a group isomorphism between these two groups.

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The model is due to Shelah, not Solovay. (Though the title of the paper is Can you take Solovay's inaccessible away?) –  François G. Dorais May 20 '10 at 17:08
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Hi François, I don't think this is entirely accurate. Solovay's model has the property Gerald mentions (every set of reals there is Lebesgue measurable, has the property of Baire, the perfect set property, is Ramsey). It is true, though, that Solovay's construction requires an inaccessible and Shelah later showed, in the paper you mentioned, that for the property of Baire one could build such a model without this assumption. –  Andres Caicedo May 20 '10 at 23:15
    
You're right Andres; I was assuming that by using category instead of measure, Gerald wanted to avoid Solovay's inaccessible. –  François G. Dorais May 21 '10 at 14:55
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